Question 25

ΔPQR is right angled at Q. If cosecP = 17/15, then what is the value of sinR ?

Solution

Given : $$\cosec P$$ = $$\frac{17}{15}$$

=> $$\sin P=\frac{QR}{PR}=\frac{15}{17}$$

Let QR = 15 cm and PR = 17 cm

Thus, in $$\triangle$$ PQR, => $$(PQ)^2=(PR)^2-(QR)^2$$

=> $$(PQ)^2=(17)^2-(15)^2$$

=> $$(PQ)^2=289-225=64$$

=> $$PQ=\sqrt{64}=8$$ cm

To find : $$\sin R=\frac{PQ}{PR}$$

= $$\frac{8}{17}$$

=> Ans - (B)

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